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- Nonspecific elution of affinity bonded macromolecules is used in affinity chromatography explain why?Can Cells Be Separated into Their Component Fractions? How? Centrifugation is the first step in most fractionations, but it separates only components that differ greatly in size. Compare and contrast very high, high, medium, and low-speed centrifugations of supernatant mixtures. What is Chromatography? Can proteins be purified and separated through Chromatography? Explain your answer. Define the following: SDS Polyacrylamide-Gel Electrophoresis and ImmunoprecipitationThinking about the complexity of biochemical systems as they relate to the human body and the specificity of DNA, why can we not describe the “average” behavior of a DNA molecule?
- TRUE OR FALSE 1. Nitrogenous bases cannot undergo electrophilic aromatic substitution.2. Cytosine forms three hydrogen bonds with guanine while Adenine forms two hydrogen bonds with Thymine in themessenger RNA.3. The glycosidic bond in nucleotides is stable in dilute ammonium hydroxide solution and dilute Sodium bicarbonatesolution.4. In gel electrophoresis, a DNA is expected to migrate towards the cathode.5. Tautomerism is a phenomenon observed in both purine and pyrimidine bases.Question: Which base (A, C, T or G) corresponds to X in the unknown? I did the experiment, got the data below, and calculated the binding constants. But I am TOTALLY lost as to how to figure this out! I don't even know what steps I would take. Base pairs Data – all had Temp = 250C PH = 7 Binding Constant A & X [A] = 0.00373221M [X] = 0.00373221M [AX] = 0.0462678M 3.322 C & X [C] = 0.0469007M [X] = 0.0469007M [CX] = 0.00309935M 1.409 T & X [T] = 0.0452279M [X] = 0.0452279M [TX] = 0.00477212M 2.333 G & X [G] = 0.0469554M [X] = 0.0469554M [GX] = 0.00304456M 2.633(b) Both laboratories used 10 micrograms of protein each in their kinetic assays. Protein concentrations weredetermined by the Bradford protein assay. Assay conditions employed in the two labs (pH, temperature,etc.) were also identical. What would be the most plausible cause for the discrepancy in the Vmax valuesfor the compound I? Explain.Recall that the Bradford assay measures total protein amounts in sample solution based on complexformation between a dye and proteins. Also, the assay solution used in both labs does not contain anyinhibitors.
- briefly explain the physical mechanism by which amino acids are separated during paper chromatography.Movement of Dyes in an Agarose Gel.This problem tests your ability to understand basic concepts of agarose gel electrophoresis.The gel is loaded from lane 1 at the top to lane 7 at the bottom with: bromophenol blue, safranin O, orange G, crystal violet, xylene cyanol, methylene blue, and the unknown.Lines have been added to the gel to help you locate the center of the wells. Which known dye samples are positively charged? Which known dyes are in the unknown sample in lane 7?Confirm Regularly bring your favourites, password... of 45 OCR 2021. You may photocopy this page. 7 of 46 Create 4 Which of the following nucleotides contains uracil? A B NH H,C. ofo HN HO-P-O- HO-P-O- OH ОН ОН OH 'NH HO-P- NH, HO-P-O OH ОН ОН OH Your answer An anticodon seguence of five successsive tRNA molecules involved in protein sunthosie was ar
- The purification continues with a cation exchange step in which the positively charged cytochrome C protein is separated from negatively charged DNA and other proteins. The cation exchange eluate (volume of solution collected) had a total volume of 42.0 mL and a 1.0 mL aliquot was set aside for further analysis. The following data was obtained from the 1.0 mL aliquot to quantify the protein amount and purity: The absorbance at 410 nm of the aliquot was diluted 5-fold was 0.474 (1 cm pathlength). The absorbance at 595 nm from a 1.0 mL Bradford Assay solution that was diluted by 100-fold from the aliquot was 0.195 (1 cm pathlength). Using the information given, Calculate the total protein amount in mg from the absorbance at 595 nm. Calculate the cytochrome C amount in mg from the absorbance at 410 nm using Beer’s Law.Could you please help me with my reaction paper. Make a simple reaction to this. Thank youFLUORESCENCE BINDING a) In the DAPI DNA lab, concentration of DAPI and DNA was determined to carry out analysis. Here is the scenario: 3.50 mL of DAPI at 25 micromolar was placed in the cuvette. It was titrated with a 150 microgram/mL DNA stock. It took 230 microliters of DNA stock to reach saturation. What is the DAPI molarity at saturation in the cuvette? What is the DNA molarity (in terms of base, 1 base = 325 g/mol) at saturation in (i) (ii) cuvette? im DNA = 150 × 10 g/mL 3.5-CAP 25 μM the 230 μL